Can a countable set contain uncountably many infinite subsets such that the symmetric difference of any two such distinct subsets is finite? There is a similar topic about finite intersections and the constructs for that case are pretty clear (for example, limits of real numbers approached by monotonically increasing sequences of rational numbers). But, is that the case for finite symmetric differences? It not obvious at all for me... What the construct can be if it is possible (is it possible to find the bijection between almost disjoint family and a family of sets with finite symmetric differences)? What's the counter-example if it's not?
 A: No, this is not possible.
Think about it this way: saying that $X\triangle Y$ is "small" is saying that $X$ is "close to" $Y$. So by counting the number of sets "close to" a given one, we can get a bound on the size of a family of sets whose pairwise symmetric differences are all "small."
Specifically, suppose $X\subseteq\mathbb{N}$ (we might as well take $\mathbb{N}$ to be our countable "base set"). Then there is a bijection between the sets close to $X$ and the finite sets of naturals - send a finite $F$ to the close-to-$X$ set $X\triangle F$, and to invert this send a close-to-$X$ set $Y$ to the finite set $X\triangle Y$ - and there are only countably many of the latter.


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*An important point here is that $\triangle$ allows cancellation: if $X\triangle A=X\triangle B$ then $A=B$. So the same finite difference can't occur more than once, and $X\triangle F$ is the only set whose symmetric difference with $X$ is $F$ (and hence the map described above is surjective).


So any collection $\mathcal{F}$ of subsets of $\mathbb{N}$ (or any countable set) whose pairwise symmetric differences are finite must be countable: picking some $X\in\mathcal{F}$, there are only countably many subsets of $\mathbb{N}$ whose symmetric difference with $X$ is finite.
